Two particles A and B, of masses 1.3 kg and 0.7 kg respectively, are connected by a light inextensible string which passes over a smooth fixed pulley. Particle A is 1.75 m above the floor and particle B is 1 m above the floor (see diagram). The system is released from rest with the string taut, and the particles move vertically. When the particles are at the same height the string breaks. [Figure X.X]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [M1] Use of Newton's Second law for A or B or use of a = (mA-mB)g/(mA+mB): Particle A: [1.3g-T=1.3a] or Particle B: [T-0.7g = 0.7a]
2. [A1] Two correct equations: 1.3g-T=1.3a and T-0.7g = 0.7a
3. [M1] Solves for a or for T
4. [A1] a = 3 ms⁻² and T= 9.1 N (a = 3)
5. [B1] Distance while connected = 0.375 m
6. [M1] Use of suvat to find v at ‘break’ (v² = 2as): [v² = 0² + 2 × 3 × 0.375 → v = ...]
7. [A1] Correct value or expression for v: v = 1.5 ms⁻¹
8. [M1] Finds one time 'from break to floor': [A: 1.375 = 1.5t + ½ gt² → t = 0.395...]
9. [M1] Finds second time 'from break to floor': [B: 1.375 = -1.5t + ½ gt² or -1.375 = 1.5t – ½ gt² → t=0.695...]
10. [A1] Difference in times = 0.3 s
11. [M1] Alternative Method 1 for 4(ii): Finds tB from 'break' to maximum height: [uB= 1.5, vB=0, a = −g, 0 = 1.5 – gt → t = 0.15]
12. [M1] Alternative Method 1 for 4(ii): Difference in times = 2 × 0.15
13. [A1] Alternative Method 1 for 4(ii): Difference in times = 0.3 s
14. [M1] Alternative Method 2 for 4(ii): Use of suvat to find total time for A: [A: 0.375 = ½ × 3t² → t = 0.5. 1.375 = 1.5t + ½ gt² → t = 0.395... tA total = 0.5 +0.395... = 0.895... s]
15. [M1] Alternative Method 2 for 4(ii): Use of suvat to find total time for B: [B: 0.375 = ½ × 3t² → t=0.5; 0 = 1.5-gt → t=0.15, s = 1.5t - 1/2gt² = 0.1125_1.4875 = ½ × gt² → t=0.545... tB total = 1.195 s]
16. [A1] Alternative Method 2 for 4(ii): Difference in times = 0.3 s